Question

If ${{a}_{1}},{{a}_{2}},{{a}_{3}}......{{a}_{n-1}},{{a}_{n}}$ are in A.P., then show that $\dfrac{1}{{{a}_{1}}{{a}_{n}}}+\dfrac{1}{{{a}_{2}}{{a}_{n-1}}}+\dfrac{1}{{{a}_{3}}{{a}_{n-2}}}............\dfrac{1}{{{a}_{n}}{{a}_{1}}}=\dfrac{2}{\left( {{a}_{1}}+{{a}_{n}} \right)}\left[ \dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{2}}}+.....\dfrac{1}{{{a}_{n}}} \right]$

Answer 1

Hint: To solve this question we need to have a knowledge of Arithmetic Progression. The series is said to be in A.P. when the common difference between the consecutive terms is the same for the terms given in the question. Fraction equal part of a whole collection.

Complete step-by-step answer:
The question ask us to prove $\dfrac{1}{{{a}_{1}}{{a}_{n}}}+\dfrac{1}{{{a}_{2}}{{a}_{n-1}}}+\dfrac{1}{{{a}_{3}}{{a}_{n-2}}}............\dfrac{1}{{{a}_{n}}{{a}_{1}}}=\dfrac{2}{\left( {{a}_{1}}+{{a}_{n}} \right)}\left[ \dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{2}}}+.....\dfrac{1}{{{a}_{n}}} \right]$, when the terms ${{a}_{1}},{{a}_{2}},{{a}_{3}}......{{a}_{n-1}},{{a}_{n}}$ are in Arithmetic Progression. The first step for the proving is to take some assumptions. The assumption that is needed to be considered is:
Let the sum of first and the last term, second and the second last term, third and the third last term be a constant $k$. On writing the above explanation in the mathematical form, we get:
$\Rightarrow {{a}_{1}}+{{a}_{n}}={{a}_{2}}+{{a}_{n-1}}={{a}_{3}}+{{a}_{n-2}}=k$
Now, considering the reciprocal of the product of the first and the last term, second and the second last term likewise we need to apply certain operation on it. Multiplying the term, $\dfrac{1}{{{a}_{1}}{{a}_{n}}}$with ${{a}_{1}}+{{a}_{n}}$ both in the numerator and the denominator, we get:
$\Rightarrow \dfrac{1}{{{a}_{1}}{{a}_{n}}}=\dfrac{{{a}_{1}}+{{a}_{n}}}{\left( {{a}_{1}}+{{a}_{n}} \right){{a}_{1}}{{a}_{n}}}=\dfrac{1}{{{a}_{1}}+{{a}_{n}}}\left( \dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{n}}} \right)$
Substituting the terms in addition with “k” we get:
$\Rightarrow \dfrac{1}{{{a}_{1}}{{a}_{n}}}=\dfrac{1}{k}\left( \dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{n}}} \right)$
So, on calculating the sum for all the terms given in the question we get:
$\dfrac{1}{{{a}_{1}}{{a}_{n}}}+\dfrac{1}{{{a}_{2}}{{a}_{n-1}}}+.........+\dfrac{1}{{{a}_{n}}{{a}_{1}}}$
Substituting the value on the above expression we get the following expression:
$\Rightarrow \dfrac{1}{k}\left( \dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{n}}} \right)+\dfrac{1}{k}\left( \dfrac{1}{{{a}_{2}}}+\dfrac{1}{{{a}_{n-1}}} \right)+........\dfrac{1}{k}\left( \dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{n}}} \right)$
Now taking the terms common, the above expression turns to be:
$\Rightarrow \dfrac{2}{k}\left( \dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{n}}}+\dfrac{1}{{{a}_{2}}}+\dfrac{1}{{{a}_{n-1}}}........ \right)$
Substituting the value of “k” with the terms above we get the following expression:
$\Rightarrow \dfrac{2}{{{a}_{1}}+{{a}_{2}}}\left( \dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{2}}}+......+\dfrac{1}{{{a}_{n}}} \right)$
$\therefore $ The expression $\dfrac{1}{{{a}_{1}}{{a}_{n}}}+\dfrac{1}{{{a}_{2}}{{a}_{n-1}}}+\dfrac{1}{{{a}_{3}}{{a}_{n-2}}}............\dfrac{1}{{{a}_{n}}{{a}_{1}}}=\dfrac{2}{\left( {{a}_{1}}+{{a}_{n}} \right)}\left[ \dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{2}}}+.....\dfrac{1}{{{a}_{n}}} \right]$ is hence proved.

Note: To check whether the series is in Arithmetic Progression or not, check the common difference. Common Difference refers to the difference between the two consecutive terms in the series.